Question

Find the minimum wavelength of X-ray produced if $10kV$ potential difference is applied across the anode and cathode of the tube.
A) $12.4\,{A^o}$
B) $12.4\,nm$
C) $1.24\,nm$
D) $1.24\,{A^o}$

Answer 1

When a potential difference is applied across the cathode and anode, the electrons will gain an acceleration and hence Kinetic Energy. When we equate this energy to wavelength in the respective formula, we can get the minimum wavelength of X-ray produced.

Formula Used:
Energy of an electron, ${E_{\max }} = \dfrac{{hc}}{{{\lambda _{\min }}}}$ where, $h = 6.6 \times {10^{ - 34}}JH{z^{ - 1}}$ is Planck’s constant, $c = 3 \times {10^8}m{s^{ - 1}}$ is the speed of light (electrons move with the speed of light), ${\lambda _{\min }}$ is the minimum wavelength of the electrons whose energy is being calculated. (when wavelength is minimum, energy will be maximum)

Complete Step by Step Solution:
When a potential difference of $10kV$ is applied across the anode and the cathode the electrons, carrying negative charge, will be attracted by the anode and hence will gain an acceleration. Since they are moving, they will also gain kinetic energy. The energy in each of the electrons will be equal to the potential difference applied. Therefore, Kinetic energy in each electron will be $10keV$
This will be the maximum energy an electron can have. Since energy and wavelength are inversely proportional, maximum energy will give us minimum wavelength.
We have, ${E_{\max }} = \dfrac{{hc}}{{{\lambda _{\min }}}}$ where, $h = 6.6 \times {10^{ - 34}}JH{z^{ - 1}}$ is Planck’s constant, $c = 3 \times {10^8}m{s^{ - 1}}$ is the speed of light, ${\lambda _{\min }}$ is the minimum wavelength of the electrons whose energy is being calculated.
We know that ${E_{\max }} = 10keV = 10000eV$ , $hc \simeq 12400eV\mathop A\limits^o$
Therefore, ${\lambda _{\min }} = \dfrac{{hc}}{{{E_{\max }}}} = \dfrac{{12400}}{{10000}} = 1.24\mathop A\limits^o$

Hence option D is the correct answer.

Note: $h$ and $c$ are constants and are widely used in modern physics. You might need to memorize their values as well as the value of their product. It may come handy in questions like these and will save you a lot of time.